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S4 Physics Waves Practice Paper 1 | v = f x lambda, Period and Echo Ranging: 10 Questions with Answers
S4 Physics Waves Practice Paper 1 | v = f x lambda, Period and Echo Ranging: 10 Questions with Answers
Ten wave questions, from v = f x lambda through period, the speed of sound, echo ranging and the frequency that stays fixed on refraction, each with an answer and a one-line explanation.
Direct answers This paper has 10 questions covering the wave speed formula, period, and sound and ultrasound ranging; about 20 minutes is suggested.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 minutes |
| Difficulty mix | Foundation 3 · Intermediate 4 · Challenge 2 · Error redo 1 |
| Chapter | S4 Physics: Waves |
| Level | S4 (DSE) |
02 Questions (10 items)
- FoundationA wave has a frequency of 50 Hz and a wavelength of 4 m. What is its speed?
- FoundationWhat is the period of a wave with a frequency of 0.5 Hz?
- FoundationHow is wavelength defined?
- IntermediateA shout towards a cliff is heard as an echo 2 s later. How far away is the cliff? (speed of sound 340 m/s)
- IntermediateA wave passes from deep water into shallow water. How do the frequency, the speed and the wavelength change?
- IntermediateWhat is the wavelength of a sound wave of frequency 250 Hz travelling at 340 m/s?
- IntermediateLight passes from air into water at an angle. How do the frequency and the speed change?
- Challenge30 complete waves pass a point in 3 s. What are the frequency and the period?
- ChallengeAn ultrasound pulse takes 0.04 s to travel to a target and back in sea water at 1 500 m/s. How far away is the target?
- Error redoIs the statement ‘sound can travel through a vacuum’ correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | v = f x lambda = 50 x 4 = 200 m/s. | Wave speed equals frequency multiplied by wavelength. |
| 2 | T = 1 / f = 1 / 0.5 = 2 s. | Period and frequency are reciprocals of each other. |
| 3 | The distance between two adjacent points in phase, for example the distance between two adjacent crests. | It is the length occupied by one complete wave cycle. |
| 4 | The sound travels 340 x 2 = 680 m, so one way is 680 / 2 = 340 m. | The echo covers the distance there and back. |
| 5 | The frequency stays the same, the speed decreases and the wavelength decreases too. | Frequency is set by the source, not by the medium. |
| 6 | lambda = v / f = 340 / 250 = 1.36 m. | Rearrange the formula before dividing, and do not multiply by mistake. |
| 7 | The frequency stays the same and the speed decreases, so the wavelength is shorter and the ray bends towards the normal. | Frequency is conserved during refraction. |
| 8 | f = 30 / 3 = 10 Hz, T = 1 / 10 = 0.1 s. | Frequency is the number of waves passing per unit time. |
| 9 | The one-way time is 0.02 s, so the distance is 1 500 x 0.02 = 30 m. | Ultrasound ranging also requires dividing by 2. |
| 10 | No. Sound is a mechanical wave and needs a medium to vibrate, so it cannot travel through a vacuum. | Without a medium there is no sound wave. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Work through all ten questions under timed conditions first, then check each answer against its one-line explanation and redo the ones you got wrong.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
